For the incoming state, you don't know which spin state the particle is in, so you should *average* over the possible states. But you can measure the spin of the outgoing state, so to get the total cross section you should *add* up the cross sections for each spin.

More formally, an unpolarized incoming particle should be described as a density matrix, $$
\def\bra#1{\mathinner{\langle{#1}|}}
\def\ket#1{\mathinner{|{#1}\rangle}}
\rho(t=-\infty) = \frac{1}{2}( \ket{p\uparrow} \bra{p\uparrow} + \ket{p\downarrow}\bra{p\downarrow})$$
where the $\frac{1}{2}$ is for the normalization of the density matrix, $\operatorname{tr} \rho = 1$. Here $p$ is the momentum and this is of course for spin $\frac{1}{2}$ but the generalization to spin $1$ and higher will be obvious.

The density matrix will evolve to $\rho(t=\infty) = S \rho(t=-\infty) S^\dagger$ just by definition of the $S$-matrix. You want to calculate the probability to end up with momentum $q$, regardless of spin, at $t=\infty$. This is the expectation value at $t=\infty$ of the projection operator $$
P(q) = P(q,\uparrow) + P(q,\downarrow) = \ket{q\uparrow}\bra{q\uparrow} + \ket{q\downarrow}\bra{q\downarrow}.
$$

The expectation value of the projection operator is $$\langle P(q) \rangle = \operatorname{tr}(\rho P(q) ).$$
There will be four terms in the trace. One of them is $$T_{\uparrow\uparrow} = \frac{1}{2}\operatorname{tr}\big(S\ket{p\uparrow}\bra{p\uparrow}S^\dagger\ket{q\uparrow}\bra{q\uparrow}\big).$$
Note that in the middle is something that is just a number, and a familiar number too. $\bra{p\uparrow}S^\dagger\ket{q\uparrow} = (\bra{q\uparrow}S\ket{p\uparrow})^*$ is (the complex conjugate) of an S-matrix element. It is easy to realize that $$\operatorname{tr}\big(S\ket{p\uparrow}\bra{q\uparrow}\big) = \bra{q\uparrow}S\ket{p\uparrow}$$ and therefore $$T_{\uparrow\uparrow} = \frac{1}{2} |\bra{q\uparrow}S\ket{p\uparrow}|^2.$$
The other terms in the trace are similar and we end up with $$\langle P(q) \rangle = \frac{1}{2} \sum_{r,s} |S(p,r;q,s)|^2$$
where $r,s \in \{\uparrow, \downarrow\}$ are incoming and outgoing spins, respectively, and $S(p,r;q,s)$ is the $S$-matrix element. As you can see we recover the prescription average over incoming, and sum over outgoing.

This post imported from StackExchange Physics at 2015-03-23 06:04 (UTC), posted by SE-user Robin Ekman